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 2022-10-31, 07:23 #1 tgan   Jul 2015 2×19 Posts November 2022
 2022-10-31, 08:25 #2 SuikaPredator   Aug 2022 China 25 Posts . Last fiddled with by SuikaPredator on 2022-10-31 at 08:26 Reason: deleted
 2022-10-31, 08:50 #3 LaurV Romulan Interpreter     "name field" Jun 2011 Thailand 3×23×149 Posts Not really. While, with the current computing power of a good CPU today, a single line in Pari incrementing the enumerator and denominator alternatively (the silly-stupid way) will give you the EP100 solution in about half hour or so (didn't have the patience and the time, stopped it after few seconds), here (by replacing 2 with the constant they give) it will not work, due to the size of the numbers, you will get quite old going one by one for 100 digits , and will have to do binary search, or come with a more clever algorithm. Code: gp > m=1; n=1; while(1, if((zn=(2*n*(n-1)))<(zm=(m*(m-1))),n++,if(zn>zm,m++,print(n", "m);n++))) 1, 1 3, 4 15, 21 85, 120 493, 697 2871, 4060 16731, 23661 97513, 137904 568345, 803761 3312555, 4684660 19306983, 27304197 112529341, 159140520 edit: Hey buddy, grrr, I just found out that you deleted the post and let mine hanging ! Last fiddled with by LaurV on 2022-10-31 at 15:02
 2022-11-01, 03:15 #4 Rubiksmath   Sep 2022 3×23 Posts a must be one of the following forms mod 487085 (i think), but considering the fact we are dealing with 100 digit numbers it hardly helps: 0,1,79850,114985,194835,292251,372101,407236. Last fiddled with by Rubiksmath on 2022-11-01 at 03:22
2022-11-01, 05:58   #5
Dieter

Oct 2017

2138 Posts

Quote:
 Originally Posted by Rubiksmath a must be one of the following forms mod 487085 (i think), but considering the fact we are dealing with 100 digit numbers it hardly helps: 0,1,79850,114985,194835,292251,372101,407236.
You wanted to say "b must be ..." ?

2022-11-01, 06:06   #6
Rubiksmath

Sep 2022

3·23 Posts

Quote:
 Originally Posted by Dieter You wanted to say "b must be ..." ?
yes you are right.

2022-11-01, 08:21   #7
SuikaPredator

Aug 2022
China

25 Posts

Quote:
 Originally Posted by LaurV edit: Hey buddy, grrr, I just found out that you deleted the post and let mine hanging !
I'm sorry, but I discovered my mistake immediately after posting that one. Actually I misread the fraction 1/974170 in the problem statement into 1/2. Maybe the actual problem could also be solved by continued fractions but with some more complex deductions than the 1/2 version.

 2022-11-02, 06:52 #8 tgan   Jul 2015 3810 Posts Please explain "Your goal: find a,b such that the probability for two comfortable socks is exactly \frac{1}{974170} , and such that this is the minimal solution with b having at least 100 digits (minimal with respect to the size of b )" My understanding is b should be bigger than 10^99 and smaller than any other potential b?
2022-11-02, 13:28   #9
Dr Sardonicus

Feb 2017
Nowhere

11000010111012 Posts

Quote:
 Originally Posted by tgan "Your goal: find a,b such that the probability for two comfortable socks is exactly \frac{1}{974170} , and such that this is the minimal solution with b having at least 100 digits (minimal with respect to the size of b )" My understanding is b should be bigger than 10^99 and smaller than any other potential b?
No, just that b is the smallest solution that is greater than or equal to 10^99.

If I did my sums correctly, the smallest possible b is 114985, with a = 117.

(I don't think this is giving away anything useful for answering the stated question.)

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